For how many values of k does the system of l | vedclass

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(B) For a system of linear equations to have no solutions,the determinant of the coefficient matrix $\Delta$ must be $0$,and at least one of the deter

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$1$

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(B) For a system of linear equations to have no solutions,the determinant of the coefficient matrix $\Delta$ must be $0$,and at least one of the determinants $\Delta_1$ or $\Delta_2$ must be non-zero.$\Delta = \begin{vmatrix} k+1 & 8 \ k & k+3 \end{vmatrix} = (k+1)(k+3) - 8k = k^2 + 4k + 3 - 8k = k^2 - 4k + 3 = (k-3)(k-1)$.Setting $\Delta = 0$,we get $k = 1$ or $k = 3$.Now,calculate $\Delta_1$ and $\Delta_2$ for these values:$\Delta_1 = \begin{vmatrix} 4k & 8 \ 3k-1 & k+3 \end{vmatrix} = 4k(k+3) - 8(3k-1) = 4k^2 + 12k - 24k + 8 = 4k^2 - 12k + 8 = 4(k-1)(k-2)$.$\Delta_2 = \begin{vmatrix} k+1 & 4k \ k & 3k-1 \end{vmatrix} = (k+1)(3k-1) - 4k^2 = 3k^2 + 2k - 1 - 4k^2 = -k^2 + 2k - 1 = -(k-1)^2$.Case $k=1$: $\Delta = 0, \Delta_1 = 0, \Delta_2 = 0$. This leads to infinitely many solutions (or consistency).Case $k=3$: $\Delta = 0, \Delta_1 = 4(3-1)(3-2) = 8 
eq 0, \Delta_2 = -(3-1)^2 = -4 
eq 0$.Since $\Delta = 0$ and $\Delta_1, \Delta_2 
eq 0$ for $k=3$,the system has no solutions.Thus,there is only $1$ value of $k$.

For how many values of $k$ does the system of linear equations $(k + 1)x + 8y = 4k$ and $kx + (k + 3)y = 3k - 1$ have no solutions?

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